Is there a way to target PHP includes only for certain pages? I have to use PHP includes to call JavaScript and CSS files only for certain pages; pages not targeted for the include will not see either the JavaScript or CSS otherwise they will break. Thanks
EDIT:
Thanks everyone. I've switched approaches. Instead of selectively using includes, I've decided to just 开发者_开发问答check the file name and link to JavaScript and CSS files to targeted pages. A colleague pointed to using$_SERVER['REQUEST_URI']
which would serve the name of the .php document and allow targeted delivery of JavaScript and CSS files. This is how I've set it up so far:
$files-array{file1.php, file2.php, file3.php}
if (array_search($_SERVER['REQUEST_URI'], $files-array) > 0) {
link to css files
link to js files
}
How is a page defined?
if($_GET['something'] == "somepage"){
include 'javascript.js';
}
You could do a bunch of things, however the main logic is:
if condition
include file
So you could do something like:
if( $_SERVER["SCRIPT_NAME"] == "folder/script.php" ){
include 'page.js';
}
or
if($_GET['variable'] == "correct"){
// include
}
or
if($_POST['formvariable'] == "correct"){
// include
}
and so on.
I'd recommend do your own library. So you don't mess the code up. You can make a simple catalog, something like this:
$pages_includes = array(
'index' => array(
'script1.js',
'script2.js',
'style1.css',
'style2.css',
),
'contact' => array(
'script1.js',
'style1.css',
'style2.css',
)
);
Then, in your "library" you could make something like this
function myImport($page_name){
$includes = $pages_includes[$page_name];
foreach(includes as $k=>$v){
include_once($v);
}
}
And, in your page you can do something like this:
myImport($_SERVER["SCRIPT_NAME"]);
Those are simple ways to organize your code and make it modular. If you have to change any imports you can do it in your Catalog ($pages_includes) and not in the code. Even you can read your catalog from a XML file.
A simple lesson of scalability and extensibility ;)
Can't you just use
if (condition) include 'something';
?
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